Algebra – Page 887

a-b-Z-b-2-a-2-2b-7a-4-a-b-

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Question Number 9962 by konen last updated on 19/Jan/17 $a, b \in Z^{+}$ $b^{2} - a^{2} = 2 b + 7 a + 4$ $\Rightarrow a + b = ?$ Commented by prakash jain last updated on…

x-2-25-4-x-5-x-

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Question Number 9959 by konen last updated on 19/Jan/17 $∣ x^{2} - 25 ∣= 4 ∣ x - 5 ∣ \Rightarrow Σ x = ?$ Answered by mrW1 last updated on 19/Jan/17 $∣ (x - 5) (x + 5) ∣= 4 ∣ x - 5 ∣$ $x = 5 o r$ $$\mid{x}+\mathrm{5}\mid=\mathrm{4}\:{or}…

n-1-1-9n-2-3n-

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Question Number 141030 by iloveisrael last updated on 15/May/21 $\to ⟨ \underset{n = 1}{\sum^{\infty}} \frac{1}{9 n^{2} + 3 n} = ? ⟩ \leftarrow$ Commented by EDWIN88 last updated on 15/May/21 $= 1 - \ln \sqrt{3} - \frac{π}{6 \sqrt{3}}$ $$\:\mathrm{S}=\frac{\mathrm{1}}{\mathrm{3}}\:\underset{\mathrm{n}=\mathrm{1}}…

x-y-8-xy-7-x-2-y-2-

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Question Number 9953 by konen last updated on 18/Jan/17 $x - y = 8$ $xy = 7 \Rightarrow x^{2} - y^{2} = ?$ Answered by mrW1 last updated on 19/Jan/17 $$\left({x}−{y}\right)^{\mathrm{2}} =\mathrm{8}^{\mathrm{2}}…

s-2-s-1-2-c-2-s-s-1-2-c-2-2-s-1-c-3-2-0-solve-for-s-in-terms-of-0-lt-c-lt-2-3-3-

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Question Number 141012 by ajfour last updated on 14/May/21 $s^{2} {(s + 1)}^{2} + \frac{c}{2} s {(s + 1)}^{2} - \frac{c^{2}}{2} (s + 1)$ $+ \frac{c^{3}}{2} = 0 s o l v e f o r s i n t e r m s$ $o f 0 < c < \frac{2}{3 \sqrt{3}} .$ Terms of Service Privacy…

Let-0-a-b-lt-1-Prove-that-1-4-2-a-2-b-1-a-1-b-4-a-b-4-a-b-

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Question Number 141004 by loveineq last updated on 14/May/21 $Let 0 ⩽ a, b < 1 . Prove that$ $\frac{1}{4} \cdot \frac{(2 - a) (2 - b)}{(1 - a) (1 - b)} ⩾ \frac{4 + a + b}{4 - a - b}$ Terms of Service Privacy Policy Contact: info@tinkutara.com

If-a-gt-0-and-one-root-of-ax-2-bx-c-0-is-less-than-2-and-the-other-is-greater-than-2-then-A-4a-2-b-c-lt-0-B-4a-2-b-c-gt-0-C-4a-2-b-c-0-D-a-b-c-

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Question Number 141002 by EnterUsername last updated on 14/May/21 $If a > 0 and one root of a x^{2} + b x + c = 0 is less than - 2$ $and the other is greater than 2, then$ $(A) 4 a + 2 ∣ b ∣ + c < 0$ $(B) 4 a + 2 ∣ b ∣ + c > 0$ $(C) 4 a + 2 ∣ b ∣ + c = 0$ $(D) a + b = c$ Answered by…

Let-a-b-c-0-Prove-that-1-8-2-a-2-b-2-c-1-a-1-b-1-c-4-a-b-c-4-a-b-c-

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Question Number 140997 by loveineq last updated on 14/May/21 $Let a, b, c ⩾ 0 . Prove that$ $\frac{1}{8} \cdot \frac{(2 + a) (2 + b) (2 + c)}{(1 + a) (1 + b) (1 + c)} ⩾ \frac{4 - a - b - c}{4 + a + b + c}$ Commented by MJS_new last updated on 15/May/21 $$\mathrm{as}\:\mathrm{easy}\:\mathrm{as}\:\mathrm{the}\:\mathrm{last}\:\mathrm{one} \…

Find-the-domain-and-range-of-relation-a-R-x-y-y-x-2-6-b-R-x-y-x-2-y-2-1-

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Question Number 75456 by peter frank last updated on 11/Dec/19 $F i n d t h e d o m a i n a n d$ $r a n g e o f r e l a t i o n$ $(a) R = {(x, y) : y = \sqrt{x^{2} - 6}}$ $(b) R = {(x, y) : x^{2} - y^{2} = 1}$ Answered by Kunal12588…

i-1-z-1-i-17-1-i-17-z-

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Question Number 9923 by konen last updated on 16/Jan/17 $i = \sqrt{- 1}$ $z = {(1 - i)}^{17} - {(1 + i)}^{17} \Rightarrow z = ?$ Commented by RasheedSoomro last updated on 16/Jan/17 $$\mathrm{z}=\:\left(\mathrm{1}−\mathrm{i}\right)^{\mathrm{17}} −\left(\mathrm{1}+\mathrm{i}\right)^{\mathrm{17}}…

Category: Algebra